Integration of $\int_{r}^{1}\rho^{2(1-\rho)}d\rho$

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I am trying to understand the solution of the following question:

For every $p>0$ and for every $0<r<1$, evaluate the integral $$I(r,p)=\iiint_{D_r}\frac{1}{(x^2+y^2+z^2)^p}dxdydz$$ Where $D$ is the space between a sphere in the center with radius $r$ and a sphere in the center with a radius of $1$. $$D_{r}=\{(x,y,z)\mid r^2\leq (x^2+y^2+z^2)^p\leq 1 \}$$

Solution: $$I(r,p) = \iiint_{D_r}\frac{1}{(x^2+y^2+z^2)^p}dxdydz = \int_{0}^{2\pi}\int_{0}^{\pi}\int_{r}^{1}\frac{1}{\rho^{2p}}\cdot \rho^2\sin(\varphi)d\rho d\varphi d\theta = 2\pi\bigg(\int_{0}^{\pi}\sin(\varphi)d\varphi\bigg)\bigg(\int_{r}^{1}\rho^{2(1-p)}d\rho\bigg) = \frac{4\pi}{3-2p}(1-r^{3-2p}), p\neq\frac{2}{3}$$

My question is about the last integration, where $$\small(1)\int_{r}^{1}\rho^{2(1-p)}d\rho=\frac{1}{3-2p}(1-r^{3-2p})$$

Can someone explain or rather show the steps on the integration of $\small(1)$?

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This is just the standard power rule $$ \int x^a dx = \frac{1}{a+1}x^{a+1}+ C$$ for $a\ne -1.$

In definite integral form: $$ \int_b^c x^adx = \frac{1}{a+1}(c^{a+1}-b^{a+1}).$$